Question about Gravitation PLEASE HELP!?
The question is.... "Each of two adjacent 2.3 kg spheres hangs froma ceiling by a string. The center to center distance of the sphere is 6.2 cm. What(small) angle does each string make with the vertical." Okay I understand that Force=GMm/r^2 is the radius here 6.2/2. and M=m=2.3. Also is the force of gravity the only force that does work here. or is there another force. The formula that I came up with was sin(theta)=1/Force gravity. however my answer was wrong. Could someone please help break down the problem for me and explain and correct me if I'm wrong! I would greatly GREATLY appreciate it. Thanks guys!
Gravitation question!?
1 N.
Help me in this gravitational question?
In the question there was an isolated sphere and it was written that i should draw lines to represent gravitational field and i did so..later in part b threre was question that a second sphere has same mass as that of isolated sphere but smaller radius.Suggest what difference,if any,there is between the patterns of field lines for the two spheres?plzz tell me if there are any differences or not and yes if there r then what r those?
Gravitation question (another)?
_____________________________________ Gravitational potential energy on surface of earth =GPE= -GMm/R where , mass of earth =M radius of earth =R Universal gravitational constant= G mass of space vehicle =m But, acceleration of gravity = g = GM/R^2 Suppose the space vehicle leaves the surface of the earth with a speed 'U' and at infinity, velocity is V Gravitational potential energy on surface of earth =GPE= -mgR (1/2)mU^2 - mgR =(1/2)mV^2 V = sq rt [ U^2 -2gR]=sq rt 43560000=6600 m/s the speed when it is infinitely far from the earth will be 6600 m/s ______________________________________...
Help with basic circular motion and gravitation questions?
Kepler's third law states that the square of the orbital period of a planet is proportional to the cube of the radius of the orbit (assuming a circular orbit....Kepler's Law is actually more general, and can be applied to eliptical orbits, in which case one uses the semimajor axis instead of the radius): P^2 = k*R^3, where k is a constant. We know that the period of the Earth's orbit is 1 year, and it's orbital radius is 1 astronomical unit (AU). If we use these units to express distance and time, then k = 1 (year^2)/(AU^3). For a planetoid that has an orbital radius of 55 AU, then, the orbital period would be: P^2 = (1 yr^2/AU^3 )*(55 AU)^3 P^2 = 166375 yr^3 P = 407.891 yr (rounds to 407.9 yr) See sources for a derivation of Kepler's 3rd law ---- A point on the outer surface of the tire has to travel a distance equal to one circumference of the tire in each revolution of the tire. If the tire is rotating at a rate of 830 revolutions per minute, then a point on the tire is traveling at a speed of 830*(circumference of the tire). The circumference of a circle is just pi times the diameter, so the speed of a point on the outer surface is traveling at a speed of: 830 revolutions/min * pi*0.58 meters/revolution = 1512.36 m/min 1512.36 m/min = 25.206 m/sec (rounds to 25.2 m/sec)
How do I solve questions of gravitation easily?
The escape velocity at a given height is [math]\sqrt {2}[/math] times the speed of the satellite in a circular orbit at the same height. The same is also true for the momentum of the satellite.Let [math]q[/math] be the momentum of the satellite in its orbit. Further you know that impulse imparted to a body is equal to change in its momentum.First case (diagram on the left):[math]P_1 = \sqrt{2}q - q[/math]In second case (diagram on the right):In order to make the resultant momentum of the satellite [math]\sqrt{2}q[/math], [math]P_2[/math] must be equal to [math]q[/math][math]P_2 = q[/math][math]\frac{P_1}{P_2} = \sqrt 2 - 1[/math]
What should be the approach to these gravitation questions?
Gravity, also called gravitation, in mechanics, the universal force of attraction acting between all matter. It is by far the weakest known force in nature and thus plays no role in determining the internal properties of everyday matter. On the other hand, through its long reach and universal action, it controls the trajectories of bodies in the solar system and elsewhere in the universe and the structures and evolution of stars, galaxies, and the whole cosmos. On Earth all bodies have a weight, or downward force of gravity, proportional to their mass, which Earth’s mass exerts on them. Gravity is measured by the acceleration that it gives to freely falling objects. AtEarth’s surface the acceleration of gravity is about 9.8 meters (32 feet) per second per second. Thus, for every second an object is in free fall, its speed increases by about 9.8 meters per second. At the surface of theMoon the acceleration of a freely falling body is about 1.6 meters per second per second.For more information please watch the below video :
Can someone help me with this gravitation problem?
In any uniform circular motion, radius (r), speed (v), and angular speed (ω) are related by: v = ωr And angular momentum, N, is N = mvr = mωr² . . . where m = mass of the body in motion. The acceleration (a) is: a = ω²r In a circular orbit, the acceleration due to circular motion, matches that due to gravity: a = GM/r² Equating the two, leads to Kepler's 3rd Law, in the form: GM = ω²r³ . . . where M = mass of the Earth If you solve that for ω, you can plug the result (which will depend on r) into the equations for N and for v to get the r-dependence of each of those. Keep in mind that GM is a constant for different bodies in Earth orbit. ω = (constant) r^(-³/₂)
Can someone help me with a gravitation problem?
Mars has a diameter of .54 times that of Earth and a mass of .11 times that of Earth. Suppose a rover was launched on Earth with the mass of 525 kg. A) How much does the rover weigh on Earth? B) How much does it weigh on Mars? i have the correct answers but can someone please show me how the problem is done? A) 5.20 x 10^3 N B) 1.00 x 10^3 N